Ammonia acts as a very good ligand but ammonium ion does not form complexes. Which is the following being a true reason?
Answer & explanation
Correct answer: option 3
The correct answer is option 3. Because \(NH_4^+\) ion does not have any lone pair of electrons.
Ammonia (\( \text{NH}_3 \)) and Ammonium Ion (\( \text{NH}_4^+ \))
Ammonia (\( \text{NH}_3 \)):
Nitrogen's Valence Electrons: Nitrogen has five valence electrons.
Bonding: Three of these electrons form single covalent bonds with three hydrogen atoms.
Lone Pair: The remaining two electrons form a lone pair on the nitrogen atom.
Ammonium Ion (\( \text{NH}_4^+ \)):
Formation: The ammonium ion is formed when ammonia (\( \text{NH}_3 \)) accepts a proton (\( \text{H}^+ \)). This protonation involves the lone pair of \( \text{NH}_3 \) forming a bond with \( \text{H}^+ \).
Bonding: In \( \text{NH}_4^+ \), nitrogen forms four single covalent bonds with four hydrogen atoms.
Lone Pair: There are no lone pairs left on the nitrogen atom in \( \text{NH}_4^+ \) because the lone pair has been used to form the fourth N-H bond.
Hybridization:
Ammonia (\( \text{NH}_3 \)):
Hybridization: The nitrogen atom in \( \text{NH}_3 \) undergoes \(sp^3\) hybridization.
Geometry: The structure is trigonal pyramidal, with the lone pair occupying one of the \(sp^3\) hybrid orbitals.
Ammonium Ion (\( \text{NH}_4^+ \)):
Hybridization: The nitrogen atom in \( \text{NH}_4^+ \) also undergoes \(sp^3\) hybridization.
Geometry: The structure is tetrahedral, with no lone pairs on the nitrogen atom.
Ability to Act as a Ligand:
Lone Pair: The presence of a lone pair on the nitrogen atom allows \( \text{NH}_3 \) to act as a ligand. It can donate this lone pair to a metal ion to form a coordinate covalent bond (also known as a dative bond).
Complex Formation: Because \( \text{NH}_3 \) can donate its lone pair, it readily forms complexes with many metal ions.
Ammonium Ion (\( \text{NH}_4^+ \)):
Lone Pair: The \( \text{NH}_4^+ \) ion does not have any lone pairs on the nitrogen atom. All valence electrons of nitrogen are involved in bonding with hydrogen atoms.
Complex Formation: Without a lone pair to donate, \( \text{NH}_4^+ \) cannot act as a ligand and thus does not form complexes with metal ions.
Correct and Incorrect Statements:
Correct Statement: Because \( \text{NH}_4^+ \) ion does not have any lone pair of electrons: This is the correct reason why \( \text{NH}_4^+ \) does not form complexes. The absence of a lone pair means there are no electrons available to donate to a metal ion.
Incorrect Statements:
Because \( \text{NH}_3 \) is a gas while \( \text{NH}_4^+ \) is in liquid form: The physical state of the species (gas vs. liquid) does not affect their ability to act as ligands. Ligand behavior is determined by electronic structure and availability of lone pairs, not by physical state.
Because \( \text{NH}_3 \) undergoes sp^3 hybridization while \( \text{NH}_4^+ \) undergoes \(sp^3d\) hybridization: Both \( \text{NH}_3 \) and \( \text{NH}_4^+ \) undergo \(sp^3\) hybridization. The difference lies in the presence or absence of a lone pair, not in the type of hybridization.
Because \( \text{NH}_4^+ \) ion has one unpaired electron while \( \text{NH}_3 \) has two unpaired electrons: This statement is incorrect. Neither \( \text{NH}_3 \) nor \( \text{NH}_4^+ \) has unpaired electrons. The critical factor is the presence of a lone pair on \( \text{NH}_3 \), which is absent in \( \text{NH}_4^+ \).
Summary: Ammonia (\( \text{NH}_3 \)) is a good ligand because it has a lone pair of electrons that can be donated to metal ions, forming coordinate covalent bonds. In contrast, the ammonium ion (\( \text{NH}_4^+ \)) does not have any lone pairs, making it unable to act as a ligand and form complexes.